In the arrangement shown in Figure P14.40, an object of mass, m = 2.0 kg, hangs

Question

In the arrangement shown in Figure P14.40, an object of mass, m = 2.0 kg, hangs from a cord around a light pulley. The length of the cord between point P and the pulley is L= 2.0 m.

(a) When the vibrator is set to a frequency of 175 Hz, a standing wave with six loops is formed. What must be the linear mass density of the cord?

(b) How many loops (if any) will result if m is changed to 18.0 kg?
loops

(c) How many loops (if any) will result if m is changed to 4.50 kg?
loops

In the arrangement shown in Figure P14.40, an object of mass, m = 2.0 kg, hangs from a cord around a light pulley. The length of the cord between point P and the pulley is L= 2.0 m. Figure P14.40 (a) When the vibrator is set to a frequency of 175 Hz, a standing wave with six loops is formed. What must be the linear mass density of the cord? (b) How many loops (if any) will result if m is changed to 18.0 kg? loops (c) How many loops (if any) will result if m is changed to 4.50 kg? loops

Explanation / Answer

so length L = separation of 7 nodes (6 loops)
L = ? + ?+ ? = 3?
or L(n) = N (?/2) >>>>>>>>>>>>>>>>
>>> N =6 >> L = 3? = 2
? = 2/3
frequency (n) = v / ? = (1/?)?(T/mo)
where T = tension = 2*g = 2*9.8 = 19.6 N
mo= lineat mass density = mass/length
n = (3/2)?(19.6/mo) = 175
a) mo = 1.44*10^-3 kg/m
------------
b) n = (1/?)?(T/mo) = (N/2L)?(mg/mo)
when m is changed from 2 kg to 18 kg (m1) >>> and n, L mo are fixed
N = k /?(m) where k is constant
N1 = k /?(m1)
N1 = N * ?(m/m1) = 2*?(2/18) = 0.666667
only 0.6666667 loop will be formed
---------------
c) N2 = N * ?(m/m2) = 2 ?(2/4.50) = 1.33334 loops

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